Couple questions about brominating amphetamine freebases.
A solution of 3,4-methylenedioxyamphetamine (MDA) in acetic acid was treated with elemental bromine, generating the hydrobromide salt of 2-bromo-4,5 methylenedioxyamphetamine"
"To a well-stirred solution of 24.8 g 2,5 dimethoxyphenethylamine in 40mL glacial acetic acid, there was added 22 g elemental bromine dissolved in 40 mL acetic acid. After a couple of min, there was the formation of solids and the simultaneous evolution of considerable
heat."
When Shulgin is brominating 2C-H or MDA in HOAc it looks like he is using 2X+ equivalents of Br, he also mentions what reads to me like a simultaneous generation of HBr which forms the HBr salt at the same time the ring is being brominated.
Do i interpret correctly? How is the HBr being formed? where is the H coming from? How can this be prevented?
he comes right and admits it here "To a well-stirred solution of 31 g 2,6-dimethoxytoluene in 200 mL CH2Cl2 there was added 11 mL elemental bromine, a portion at a
time. There was a copious evolution of HBr and the color gradually faded from deep red to straw."
Yet in beakers synth he is only generating 1.1 equivalents of Br.
"43.8g (94%) of 2CH Hydrochloride basifiied , which was dissolved in 500 mL of 3:1 AcOH/H2O. The rxn was cooled to 0C in an ice/water bath. 37.3g of 48% aq. HBr was added, followed immediately by 23.8g of 30% H2O2"
Why does his variation not require the extra equivalent of bromine? Is no HBr being generated in this rxn? Is the ring brominated before HBr salt can be formed, or does the water present inhibit the HBr generation?
Osmium mentioned brominating in aqueous HBr, and all those vanillin bromination recipes are only using 1.1 equivalents, no HBr generation there either
Last question, any reason to think these methods would fail to brominate the 6 position in an otherwise ring-saturated amphetamine?
thanks in advance